Monthly Mathematics: Group theory and Topology

This is an idea that popped into my mind a while back that I’ve really wanted to make a thing. I think the monthly mathematics could be a really useful facility for skeptics to practice critical thinking skills and logic. As a skeptic, I believe critical thinking is a corner stone of what it means to be a skeptic. We don’t accept things on face value, and, ideally, question our most core of values to ensure accuracy. What does it mean to think critically though? Critical thinking, to me, seems to be an umbrella term for applying various forms of rigor in our evaluation of questions and statements about various topics. I think logic is at the core of all of the forms of rigor we apply. Mathematics is, in essence, applied logic. It is for these reasons that I am inclined to believe that practicing mathematics can benefit skeptics. Also, math is objectively—yes, objectively—fun!

Monthly Mathematics post’s aim is to facilitate such practice. In each post, I will provide math problems of various difficulties that will require a proof; this will not be a place for simple calculation, and solving the problems might take some time. Before the questions are posed, I will provide relevant definitions and theorems in hopes of providing the basic framework for thinking about the problem. These questions will not come out of a book; they will, instead, try to force the reader to connect various concepts from different fields within mathematics. The questions might be true or false, and I will provide the answers to the problems in the next post with the names of the people who got a question right first. I will also be open to suggestions for problems. Suggestions would actually be very helpful because I am a pure mathematician, and don’t have much training or interest in applied mathematics.

Don’t be afraid to be wrong! There is no shame in it, and you might even inspire someone else to get the right answer. For example, last night I gave a 2 hour talk in a hyperspace seminar, and one of my proofs was wrong. There was a gab in my argument, but I was close enough to the answer that we were able to fix it on the fly. No one shamed or laughed at me, and I didn’t let it throw me off. It was a learning exercise, and I won’t make a similar mistake again, hopefully. So, I encourage you to post your thoughts even if they aren’t completely correct or finished.

With the preliminaries out of the way, let’s give this a go. This month we will be looking at a mix of basic group theory and topology.

I have some strong opinions on mathematical notation; accordingly, my notation isn’t always standard. I will clarify notation in each post that isn’t standard, or that may be unclear. The only non-standard bit of notation is this post is my symbol for “such that:” |. I use the pipe because I find it to be more clear and consistent. I also use a slightly different notation within the conditions of a set. It is as follows: ( statement )[ conditions ], and is read as “statement such that conditions.”


Topological space: A topological space (X, T) is an ordered pair where X is some set, and where T is a set of subsets of X that meet the following conditions:
1. X \in T and \emptyset \in T
2. T is closed with respect to finite and infinite unions, i.e. if S \subset T, then [\cup\{ V | V \in S \}] \in T
3. T is closed with respect to finite intersections, i.e \forall V, U \in T, V \cap U \in T.

Open set: U is said to be an open set of the topological space iff U \in T
Point: x is said to be a point of the topological space iff p \in X
Base for a topology: The statement that \mathcal{B}, where \mathcal{B} \subset T , is a basis for (X, T) means that if  U \in T and p \in U, then \exists V \in \mathcal{B} | p \in V and V \subseteq U.
Group: A group is a set G paired with a binary operation, *, such that the following properties hold with respect to the operation
1. Closure: \forall a, b \in G, a * b \in G
2. Associativity: \forall a, b, c \in G, a * (b * c) = (a * b) * c
3. Identify element: \exists e \in G | \forall a \in G, a * e = a
4. inverse elements: \forall a \in G, \exists a^{-1} | a * a^{-1} = e

Subgroup: Let G be some group, and U \subseteq G. U is a subgroup of G iff U is a group with respect to the same operation as G.
Normal subgroup: Let G be some group, and N \subseteq G. N is said to be a normal subgroup of G iff \forall n \in N, \forall g \in G, (g)(n)(g^-1) \in N—note this is normal multiplication, cosets to be exact, here, and not the operation defined on the group.

1. If (X, T) is a topological space and \mathcal{B} is a basis for T, then \forall U \in T, \exists A \subseteq \mathcal{B} | U = \cup \{V | V \in A \} .
2. Let X be a set and \mathcal{B} be a collection of subsets of X. \mathcal{B} is a basis for a topology on X iff the following conditions hold:

  • X = \cup\{ A | A \in \mathcal{B} \}
  • if U and  V are sets in \mathcal{B} , and p \in U \cap V, then \exists W \in B | p \in W , W \subseteq U \cap V.


  1. Let G be some finite group. Does the set of normal subgroups form a topology?

  2. Does the set of normal subgroups, for some finite group G, form a base for a topology?

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The Queer Pastry

Belle is a red bean paste filled androsexual, mildly closeted transwoman pastry who is working towards becoming a mathematician. She tends to spend her time alone with a glass of wine, and rarely goes out except for classes and the occasional swordplay. While she does have a very close network of friends, she could be described as a hermit of sorts. She also feels strange about talking about herself in the third person. Can you guess what pastry she is?


  1. March 1, 2013 at 2:26 pm —

    Love the idea. As someone with a PhD in math, should I be holding my answers back so that other people have a chance to answer?

  2. March 1, 2013 at 3:08 pm —

    1. The empty set technically isn’t a normal subgroup, since it’s not a group. However, even adding the empty set won’t make the set of normal subgroups a topology, because it’s not closed under unions. For example, in the Klein four-group, {1,a} and {1,b} are normal subgroups, but {1,a,b} is not even a group.
    2. We can prove that the set of normal subgroups is a topology by theorem 2. First, note that every group is a normal subgroup of itself, fulfilling the first condition. Second, normal subgroups are closed under intersections. (I’m skipping a few steps here ’cause I’m not sure if I can write math notation in a comment.) I can’t really visualize what kind of topology it is though. Clearly not a hausdorff space.

    It’s been a few years since I took topology or group theory (I minored in math), and I never thought to combine the two in this way.

    • March 1, 2013 at 3:29 pm —

      1. That’s correct.

      2. I think you meant a base for a topology. You are correct here as well, and I am unsure what the topology would look like either, haha. Also, WordPress supports latex, and I think it supports it for comments… let’s find out A \cap B = \emptyset

      Well, I think I need to have some more hard problems mixed in. I didn’t throw any in because I didn’t want to scare everyone off, and I wasn’t sure how many math people read Queereka.

      • March 1, 2013 at 5:05 pm —

        My first reaction to this post was, “Yay, math on Queereka!” My second reaction was, “Wait, this is full on topology, with \LaTeX and everything. Won’t that scare everyone off?” But it doesn’t scare me off, so what do I know.

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